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I came across this problem:

What is the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice is i, i = 2, 3, ... , 12?

I know the answer is zero for i<7 (obviously) and I counted up the different possibilities to find the answers for i>=7, which is (1/3,2/5,1/2,2/3,1,1) for (7,8,9,10,11,12). However, I tried to deduce this with conditional probability rules, and I'm getting different answers.

Given R is event that at least one of the die lands on a six, and Si is the event that the sum of the die is i, then

P(R∣Si) = P(RSi)/P(Si) = P(R)P(Si|R)/P(Si).

P(Si|R) should be 1/6 for any value i>6, since if we know one die gets a 6, then the other die must be a specific value to get any specific value of i.

Since R is the odds that the first die rolled lands on a 6 or that the second die rolled lands on a 6, then P(R) = P(Die 1 is 6 or Die 2 is 6) = P(Die 1 is 6) + P(Die 2 is 6) - P(Die 1 is 6 and Die 2 is 6) = (1/6) + (1/6) - (1/36) = 11/36

However, if I try to use the equation and these values, I get different answers than I should. For example, for i = 12 I get

((11/36)*(1/6))/(1/36) = 11/6

Which is of course wrong because a probability can't be greater than 1. What am I doing wrong?

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    Not following your calculation...the lack of formatting is a problem. If the sum is $12$ then there is only one combination that yields $12$ (i.e. $(6,6)$) so the answer is $1$ in this case. Same if the roll is $11$ (as the only possible combinations are $(5,6),(6,5)$ and both contain a $6$). If the roll is $10$ we have three equi-probable rolls and two of those have a $6$ so $\frac 23$ in this case. And so on. – 2017-02-27

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You have an error in calculating $Pr(S_{12}\mid R)$. The probability the sum is $12$ given the first die is a 6 is indeed $\frac{1}{6}$ but this is different than the probability the sum is $12$ given at least one of the dice is a 6. There are $11$ equally likely scenarios in which at least one die is a six and exactly one of those scenarios results in the sum being twelve.

You have instead $Pr(S_{12}\mid R)=\frac{1}{11}$ not $\frac{1}{6}$ resulting in $P(R\mid S_{12})=P(R)P(S_{12}\mid R)/P(S_{12})=(\frac{11}{36}\cdot \frac{1}{11})/(\frac{1}{36})=1$ as expected.

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    Thank you, I suspected my calculation for P(S12|R) was the problem, but I couldn't tell why is was. – 2017-02-27